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Bad Student · « **on:** January 25, 2013, 11:42:16 pm »

Could someone please help me with this question?

The series $S_n$ is given by $S_n=2n^2+14n$ . The $n^{th}$ term, $t_n$ is equal to:

A 4n + 12
B 4n - 12
C 2n + 7
D 2n - 7
E 2n + 14

Limista · « **Reply #1 on:** January 25, 2013, 11:46:59 pm »

I don't think sequences and series is on the study design for spesh?

Bad Student · « **Reply #2 on:** January 25, 2013, 11:55:23 pm »

Are you sure?

abcdqd · « **Reply #3 on:** January 25, 2013, 11:56:14 pm »

certain

polar · « **Reply #4 on:** January 26, 2013, 11:25:31 am »

actually, from the study design:

Quote

Familiarity with sequence and series notation and related simple applications, the use of sine and cosine rules in non-right-angled triangles and the following mathematics is assumed

first, note that $t_1=S_1=2+14=16$

$\begin{aligned} S_n &= \frac{n}{2}(t_1 + t_n)<br />\\ 2n^2 + 14n &= \frac{n}{2}(t_1+t_n)<br />\\ t_1 + t_n &= \frac{2}{n}(2n^2+14n)<br />\\ t_n &= (4n+28) - 16<br />\\ &= 4n + 12 \end{aligned}$

fred42 · « **Reply #5 on:** January 26, 2013, 08:39:15 pm »

Please see attached.

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